C# 获取十进制数的整数部分的最佳方法
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Best way to get whole number part of a Decimal number
提问by Yaakov Ellis
What is the best way to return the whole number part of a decimal (in c#)? (This has to work for very large numbers that may not fit into an int).
返回小数的整数部分的最佳方法是什么（在 C# 中）？（这必须适用于可能不适合 int 的非常大的数字）。
GetIntPart(343564564.4342) >> 343564564
GetIntPart(-323489.32) >> -323489
GetIntPart(324) >> 324
The purpose of this is: I am inserting into a decimal (30,4) field in the db, and want to ensure that I do not try to insert a number than is too long for the field. Determining the length of the whole number part of the decimal is part of this operation.
这样做的目的是：我要插入数据库中的十进制 (30,4) 字段，并希望确保我不会尝试插入比该字段过长的数字。确定小数的整数部分的长度是此操作的一部分。
采纳答案by Yaakov Ellis
By the way guys, (int)Decimal.MaxValue will overflow. You can't get the "int" part of a decimal because the decimal is too friggen big to put in the int box. Just checked... its even too big for a long (Int64).
顺便说一句， (int)Decimal.MaxValue 会溢出。您无法获得小数的“int”部分，因为小数太大而无法放入 int 框中。刚刚检查过......它甚至太大了很长时间（Int64）。
If you want the bit of a Decimal value to the LEFT of the dot, you need to do this:
如果您想要点的左侧的 Decimal 值的位，您需要执行以下操作：
Math.Truncate(number)
and return the value as... A DECIMAL or a DOUBLE.
并将值返回为...一个 DECIMAL 或一个 DOUBLE。
edit: Truncate is definitely the correct function!
编辑：截断绝对是正确的功能！
回答by Noldorin
You just need to cast it, as such:
您只需要投射它，如下所示：
int intPart = (int)343564564.4342
If you still want to use it as a decimal in later calculations, then Math.Truncate (or possibly Math.Floor if you want a certain behaviour for negative numbers) is the function you want.
如果您仍想在以后的计算中将其用作小数，那么 Math.Truncate（或者可能是 Math.Floor，如果您想要负数的某种行为）就是您想要的函数。
回答by Keith
Depends on what you're doing.
取决于你在做什么。
For instance:
例如：
//bankers' rounding - midpoint goes to nearest even
GetIntPart(2.5) >> 2
GetIntPart(5.5) >> 6
GetIntPart(-6.5) >> -6
or
或者
//arithmetic rounding - midpoint goes away from zero
GetIntPart(2.5) >> 3
GetIntPart(5.5) >> 6
GetIntPart(-6.5) >> -7
The default is always the former, which can be a surprise but makes very good sense.
默认值始终是前者，这可能令人惊讶，但非常有意义。
Your explicit cast will do:
你的明确演员会做：
int intPart = (int)343564564.5
// intPart will be 343564564
int intPart = (int)343564565.5
// intPart will be 343564566
From the way you've worded the question it sounds like this isn't what you want - you want to floor it every time.
从你提出问题的方式来看，这听起来不是你想要的——你每次都想把它写下来。
I would do:
我会做：
Math.Floor(Math.Abs(number));
Also check the size of your decimal- they can be quite big, so you may need to use a long.
还要检查您的大小decimal- 它们可能很大，因此您可能需要使用long.
回答by Mark Carpenter
I think System.Math.Truncateis what you're looking for.
我认为System.Math.Truncate就是你要找的。
回答by Amit Gohel
Very easy way to separate value and its fractional part value.
分离值及其小数部分值的非常简单的方法。
double d = 3.5;
int i = (int)d;
string s = d.ToString();
s = s.Replace(i + ".", "");
s is fractional part = 5 and
i is value as integer = 3
s 是小数部分 = 5 并且
i 是整数值 = 3
回答by Mattheu Norwood
Public Function getWholeNumber(number As Decimal) As Integer
Dim round = Math.Round(number, 0)
If round > number Then
Return round - 1
Else
Return round
End If
End Function
回答by luka
I hope help you.
我希望能帮到你。
/// <summary>
/// Get the integer part of any decimal number passed trough a string
/// </summary>
/// <param name="decimalNumber">String passed</param>
/// <returns>teh integer part , 0 in case of error</returns>
private int GetIntPart(String decimalNumber)
{
if(!Decimal.TryParse(decimalNumber, NumberStyles.Any , new CultureInfo("en-US"), out decimal dn))
{
MessageBox.Show("String " + decimalNumber + " is not in corret format", "GetIntPart", MessageBoxButtons.OK, MessageBoxIcon.Error);
return default(int);
}
return Convert.ToInt32(Decimal.Truncate(dn));
}
